Select Chapter Topics:
Q. No.For a given cell, a 0.1 molar solution has a resistance of \(20 \ \Omega\) and molar conductivity of \(0.154 \times 10^{-3} S~cm^2~mol^{-1} \).
The value of the cell constant is:
1.
\(3.08 \times 10^{-7} cm^{-1}\)
2.
\(30.8 \times 10^{-7} cm^{-1}\)
3.
\(0.308 \times 10^{-9} cm^{-1}\)
4.
\(4.08 \times 10^{-6} cm^{-1}\)
Subtopic: Conductance & Conductivity |
78%
Level 2: 60%+
JEE
Please attempt this question first.
Hints
Please attempt this question first.
What is the nearest integer value of \(x\) for the Gibbs free energy change at 298 K, expressed as \(x×10^{−1}~kJ mol^{−1}\) for the given reaction.
\(Cu(s)+Sn^{2+}(0.001 M)\rightarrow Cu^{2+}(0.01M)+Sn(s) \)
Given: \(\small{E^\ominus_{Cu^{2+}/Cu}=0.34~V;~E^\ominus_{Sn^{2+}/Sn}=-0.14~V;~F=96500~C~mol^{-1}}\)
1. 873
2. 983
3. 1002
4. 911
Subtopic: Relation between Emf, G, Kc & pH |
56%
Level 3: 35%-60%
JEE
Please attempt this question first.
Hints
Please attempt this question first.
The limiting molar conductivities of Nal, NaNO3, and AgNO3 are \(\text { 12. } 7 \mathrm{~Sm}^2 \mathrm{~mole}^{-2}, 12 \mathrm{~Sm}^2 \mathrm{~mole}^{-2}\) and \(\text { 13.3 } \mathrm{Sm}^2 \mathrm{~mole}^{-2}\) respectively (all at 25°C). The limiting molar conductivity of Agl at this temperature is:
1. \(14~S~m^2mole^{-2}\)
2. \(16~S~m^2mole^{-2}\)
3. \(48~S~m^2mole^{-2}\)
4. \(12.6~S~m^2mole^{-2}\)
1. \(14~S~m^2mole^{-2}\)
2. \(16~S~m^2mole^{-2}\)
3. \(48~S~m^2mole^{-2}\)
4. \(12.6~S~m^2mole^{-2}\)
Subtopic: Kohlrausch Law & Cell Constant |
91%
Level 1: 80%+
JEE
Please attempt this question first.
Hints
Please attempt this question first.
How many elements can liberate H2 from dilute acids?
V, Cr, Mn, Fe, Co, Ni, Cu
1. 5
2. 4
3. 2
4. 6
V, Cr, Mn, Fe, Co, Ni, Cu
1. 5
2. 4
3. 2
4. 6
Subtopic: Electrochemical Series |
55%
Level 3: 35%-60%
JEE
Please attempt this question first.
Hints
Please attempt this question first.
Consider the following cell representation:
Pt/H2(1 atm)/H+(1 M) || Fe+3/Fe+2
The ratio of the concentration of Fe+2 to Fe+3 is-
[Given Ecell = 0.712, E0cell = 0.771]
1. 12
2. 10
3. 16
4. 8
Pt/H2(1 atm)/H+(1 M) || Fe+3/Fe+2
The ratio of the concentration of Fe+2 to Fe+3 is-
[Given Ecell = 0.712, E0cell = 0.771]
1. 12
2. 10
3. 16
4. 8
Subtopic: Nernst Equation |
85%
Level 1: 80%+
JEE
Please attempt this question first.
Hints
Please attempt this question first.
For a given reaction in an acidic medium,
\(\small{5 e^{-}+8 H^{+} \ + \ Mn O_4^{-}(0.001 \ M) \rightarrow \mathrm{Mn}^{2+}(0.1 \ M) \ + \ 4 \mathrm{H}_2 \mathrm{O}} \)
(Given values are: \(E_{cell}^o \) = 1.54 V, \(E_{cell} \) = 1.2832 V)
The \(pH\) of the solution is:
1. 2.46
2. 3.52
3. 6.38
4. 1.02
\(\small{5 e^{-}+8 H^{+} \ + \ Mn O_4^{-}(0.001 \ M) \rightarrow \mathrm{Mn}^{2+}(0.1 \ M) \ + \ 4 \mathrm{H}_2 \mathrm{O}} \)
(Given values are: \(E_{cell}^o \) = 1.54 V, \(E_{cell} \) = 1.2832 V)
The \(pH\) of the solution is:
1. 2.46
2. 3.52
3. 6.38
4. 1.02
Subtopic: Relation between Emf, G, Kc & pH |
57%
Level 3: 35%-60%
Please attempt this question first.
Hints
Please attempt this question first.
For a given cell at T K,
\(Pt/H_2 (g)(1 \ bar)/H^+(1 \ M) \ || \ Fe^{3+} /Fe^{2+}/Pt \)
E cell = 0.712 V
E0 cell = 0.770 V
If \([Fe^{2+}] \over [Fe^{3+}]\) is t, then the value of t in the expression \(({t \over 5})\) is:
(Given: \(({2.303 ~RT \over F} = 0.058) \))
1. 4
2. 6
3. 2
4. 1
\(Pt/H_2 (g)(1 \ bar)/H^+(1 \ M) \ || \ Fe^{3+} /Fe^{2+}/Pt \)
E cell = 0.712 V
E0 cell = 0.770 V
If \([Fe^{2+}] \over [Fe^{3+}]\) is t, then the value of t in the expression \(({t \over 5})\) is:
(Given: \(({2.303 ~RT \over F} = 0.058) \))
1. 4
2. 6
3. 2
4. 1
Subtopic: Nernst Equation |
75%
Level 2: 60%+
JEE
Please attempt this question first.
Hints
Please attempt this question first.
How many moles of electrons are required to reduce 1 mole of permanganate ions into manganese dioxide?
1. 3
2. 6
3. 4
4. 7
1. 3
2. 6
3. 4
4. 7
Subtopic: Faraday’s Law of Electrolysis |
83%
Level 1: 80%+
JEE
Please attempt this question first.
Hints
Please attempt this question first.
The resistivity of a 0.8 M solution of an electrolyte is 5×10−3 Ω cm. If λm is 2.5×10x,
the value of x is:
1. 4
2. 8
3. 5
4. 9
the value of x is:
1. 4
2. 8
3. 5
4. 9
Subtopic: Conductance & Conductivity |
79%
Level 2: 60%+
JEE
Please attempt this question first.
Hints
Please attempt this question first.
For a cell reaction involving a two-electron change, the standard Emf of the cell is found to be 0.295V at 25oC . The equilibrium constant of the reaction at 25oC will be:
| 1. | \(1 \times 10^{-10}\) | 2. | \(29.5 \times 10^{-2}\) |
| 3. | \(10\) | 4. | \(1 \times 10^{10}\) |
Subtopic: Relation between Emf, G, Kc & pH |
63%
Level 2: 60%+
JEE
Please attempt this question first.
Hints
Please attempt this question first.
Select Chapter Topics:
© 2026 GoodEd Technologies Pvt. Ltd.